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julsineya [31]
2 years ago
12

SQ is an angle bisector find x and the perimeter of the triangle PRQ 32,40,2x+3, 3x-1

Mathematics
1 answer:
Ket [755]2 years ago
8 0
Yes he and his family have a good day tomorrow and I will let you the mason jar of the money he is going we I am so sorry I just need points for the questions
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What value of k makes the equation true? Help plz plz
vodomira [7]

Answer:

k=15

Step-by-step explanation:

8 0
2 years ago
A diagonal of a cube goes from one of rhe cube's top corners to the opposite corner of the base of the cube. Find the length of
miss Akunina [59]
So we are trying to find this red line's length.

We can either find it directly, or use the blue line firs, and then use it as a leg for the green triangle.

So the blue leg is a hypotenuse for two of the edges. So:

blue^2 = leg^2 + leg^2 from the Pythagorean Theorem
OR
blue =  \sqrt{leg^2 + leg^2}

Which works out to:

blue =  \sqrt{10^2 + 10^2} =  \sqrt{100+100} = \sqrt{200} =  \sqrt{(100)(2)}=10 \sqrt{2}

So now that we have that, using the Pythagorean Theorem again gives:

red =  \sqrt{blue^2 + 10^2} =  \sqrt{(10 \sqrt{2})^2+10^2}= \sqrt{200+100}= \sqrt{300}
\sqrt{300}= \sqrt{100*3}=10 \sqrt{3}

So the length of the red line is found that way.

But wait!  There's more!

As it turns out, the red line can be found with an easier way that works with cubes and boxes (cuboids). It's really easy:

a^2 + b^2+c^2=d^2

Where a, b, and c are all 10m, and d is the red line. This greatly reduces the math:

d =  \sqrt{10^2+10^2+10^2} = \sqrt{100+100+100} =  \sqrt{300}

which gives the same answer as above, which you can see.

3 0
3 years ago
Connor earns $24 working every 1.5 hours.<br> Complete the table using equivalent ratios.
adelina 88 [10]
Sorry but where is the table my screen does not show it
3 0
3 years ago
Read 2 more answers
Step by step how to solve this equation by completing the square<br> x^2-16x+-60
anygoal [31]

Answer:

Step-by-step explanation:

x^2-16x+60

(x-6)(x-10) factor the equation

x-6=0

x=6

x-10=0

x=10

5 0
3 years ago
Are these equations infinite solution, no solution, or one solution?
Karo-lina-s [1.5K]

Answer:

1) one solution

2) Infinitely many solution

3) no solution

Step-by-step explanation:

The first equation is -14-8x=-2(-3x+7)

Let us expand the right hand side to get:

-14-8x=-2*-3x+-2*7

We multiply -14-8x=6x-14

We group similar terms to get:

-8x-6x=-14+14

Combine similar terms to get:

-14x=0

Divide through by -14 to get:

x=0

2) The second equation is 3+5x=5(x+2)-7

We expand to obtain:

3+5x=5x+5*2-7

Multiply to get:

3+5x=5x+10-7

This gives us

3+5x=5x+3

5x=5x

1=1

There is infinitely many solutions

3) The third equation is: 36-7x=-7(x-5)

We expand to get:

36-7x=-7x+35

Group similar terms;

-7x+7x=35-36

0=-1

This statement is not true so there is no solution.

5 0
3 years ago
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